Question -
Answer -
Given:
P (n) = n2 + n is even and P (r) is true,then r2 + r is even
Let us consider r2 + r = 2k … (i)
Now, (r + 1)2 + (r + 1)
r2 + 1 + 2r + r + 1
(r2 + r) + 2r + 2
2k + 2r + 2 [from equation (i)]
2(k + r + 1)
2μ
∴ (r + 1)2 + (r + 1) is Even.
Hence, P (r + 1) is true.